CC_Azusa

315 reputation

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visits	member for	1 year, 6 months
	seen	Apr 18 at 22:50
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I'm a physics graduate student at the University of Tokyo

	315 reputation	bio	website		visits	member for	1 year, 6 months
16 badges		location			seen	Apr 18 at 22:50

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Nov 23	accepted	A Galois Group Problem
Nov 20	accepted	Bounding $H_{0}^{1}$ norm
Nov 19	awarded	Yearling
Nov 19	comment	Bounding $H_{0}^{1}$ norm @Jose27 I see. I believe $a^{ij}$ should be symmetric, and belong to $L^{\infty}$.
Nov 19	comment	Bounding $H_{0}^{1}$ norm @Jose27 I don't think Poincare's inequality would work since that requires $1\leq p < n$. In this case, $p=2$, but we don't know anything about $n$.
Nov 19	asked	Bounding $H_{0}^{1}$ norm
Oct 19	comment	The set of real points of a variety $V$ is dense in $V$ @MattE Thanks Matt, that makes more sense
Oct 19	comment	The set of real points of a variety $V$ is dense in $V$ @MattE I'm currently seraching notations from the book "Algebraic Curves, Algebraic Manifolds and Schemes" by Danilov and Shokurov.
Oct 19	comment	The set of real points of a variety $V$ is dense in $V$ @QiaochuYuan In fact, we don't have such definition in our lecture note (We always have HW assignments like this that contain unknown notations..). However, I guess since $V$ is determined by a prime ideal $P$, does $dim_R(V(R))$ means the kul dimension of $P$ in the ring $\mathbb{R}[x_1,..,x_n]$?
Oct 19	revised	The set of real points of a variety $V$ is dense in $V$ deleted 419 characters in body
Oct 19	comment	The set of real points of a variety $V$ is dense in $V$ Oh yes.. I think I misunderstood the definition of dimension of an algebraic variety.
Oct 19	asked	The set of real points of a variety $V$ is dense in $V$
May 15	awarded	Tumbleweed
May 11	accepted	Show the norm map is surjective
May 11	comment	Show the norm map is surjective @DylanMoreland Yeah. I'm stucking here. I believe it couldn't be larger, but I can't see it.
May 11	revised	Show the norm map is surjective added 14 characters in body
May 11	asked	Show the norm map is surjective
May 10	comment	Integrally closed with roots of identity Very smart way to construct the minimal polynomial for $\Lambda$
May 10	accepted	Integrally closed with roots of identity
May 10	awarded	Commentator